Certification Problem
Input (TPDB TRS_Standard/Zantema_05/z27)
The rewrite relation of the following TRS is considered.
f(0,1,x) |
→ |
f(g(x),g(x),x) |
(1) |
f(g(x),y,z) |
→ |
g(f(x,y,z)) |
(2) |
f(x,g(y),z) |
→ |
g(f(x,y,z)) |
(3) |
f(x,y,g(z)) |
→ |
g(f(x,y,z)) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
f#(0,1,x) |
→ |
f#(g(x),g(x),x) |
(5) |
f#(g(x),y,z) |
→ |
f#(x,y,z) |
(6) |
f#(x,g(y),z) |
→ |
f#(x,y,z) |
(7) |
f#(x,y,g(z)) |
→ |
f#(x,y,z) |
(8) |
1.1 Reduction Pair Processor with Usable Rules
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) |
= |
1 |
|
weight(g) |
= |
1 |
|
|
|
in combination with the following argument filter
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
f#(x,y,g(z)) |
→ |
f#(x,y,z) |
(8) |
could be deleted.
1.1.1 Semantic Labeling Processor
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,1}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 2):
[f(x1, x2, x3)] |
= |
0 |
[0] |
= |
0 |
[1] |
= |
1 |
[f#(x1, x2, x3)] |
= |
0 |
[g(x1)] |
= |
1x1
|
We obtain the set of labeled pairs
f#010(0,1,x) |
→ |
f#000(g0(x),g0(x),x) |
(9) |
f#000(g0(x),y,z) |
→ |
f#000(x,y,z) |
(10) |
f#001(g0(x),y,z) |
→ |
f#001(x,y,z) |
(11) |
f#010(g0(x),y,z) |
→ |
f#010(x,y,z) |
(12) |
f#011(g0(x),y,z) |
→ |
f#011(x,y,z) |
(13) |
f#100(g1(x),y,z) |
→ |
f#100(x,y,z) |
(14) |
f#101(g1(x),y,z) |
→ |
f#101(x,y,z) |
(15) |
f#110(g1(x),y,z) |
→ |
f#110(x,y,z) |
(16) |
f#111(g1(x),y,z) |
→ |
f#111(x,y,z) |
(17) |
f#011(0,1,x) |
→ |
f#111(g1(x),g1(x),x) |
(18) |
f#000(x,g0(y),z) |
→ |
f#000(x,y,z) |
(19) |
f#001(x,g0(y),z) |
→ |
f#001(x,y,z) |
(20) |
f#010(x,g1(y),z) |
→ |
f#010(x,y,z) |
(21) |
f#011(x,g1(y),z) |
→ |
f#011(x,y,z) |
(22) |
f#100(x,g0(y),z) |
→ |
f#100(x,y,z) |
(23) |
f#101(x,g0(y),z) |
→ |
f#101(x,y,z) |
(24) |
f#110(x,g1(y),z) |
→ |
f#110(x,y,z) |
(25) |
f#111(x,g1(y),z) |
→ |
f#111(x,y,z) |
(26) |
and the set of labeled rules:
f010(0,1,x) |
→ |
f000(g0(x),g0(x),x) |
(27) |
f011(0,1,x) |
→ |
f111(g1(x),g1(x),x) |
(28) |
f000(g0(x),y,z) |
→ |
g0(f000(x,y,z)) |
(29) |
f001(g0(x),y,z) |
→ |
g0(f001(x,y,z)) |
(30) |
f010(g0(x),y,z) |
→ |
g0(f010(x,y,z)) |
(31) |
f011(g0(x),y,z) |
→ |
g0(f011(x,y,z)) |
(32) |
f100(g1(x),y,z) |
→ |
g0(f100(x,y,z)) |
(33) |
f101(g1(x),y,z) |
→ |
g0(f101(x,y,z)) |
(34) |
f110(g1(x),y,z) |
→ |
g0(f110(x,y,z)) |
(35) |
f111(g1(x),y,z) |
→ |
g0(f111(x,y,z)) |
(36) |
f000(x,g0(y),z) |
→ |
g0(f000(x,y,z)) |
(37) |
f001(x,g0(y),z) |
→ |
g0(f001(x,y,z)) |
(38) |
f010(x,g1(y),z) |
→ |
g0(f010(x,y,z)) |
(39) |
f011(x,g1(y),z) |
→ |
g0(f011(x,y,z)) |
(40) |
f100(x,g0(y),z) |
→ |
g0(f100(x,y,z)) |
(41) |
f101(x,g0(y),z) |
→ |
g0(f101(x,y,z)) |
(42) |
f110(x,g1(y),z) |
→ |
g0(f110(x,y,z)) |
(43) |
f111(x,g1(y),z) |
→ |
g0(f111(x,y,z)) |
(44) |
f000(x,y,g0(z)) |
→ |
g0(f000(x,y,z)) |
(45) |
f001(x,y,g1(z)) |
→ |
g0(f001(x,y,z)) |
(46) |
f010(x,y,g0(z)) |
→ |
g0(f010(x,y,z)) |
(47) |
f011(x,y,g1(z)) |
→ |
g0(f011(x,y,z)) |
(48) |
f100(x,y,g0(z)) |
→ |
g0(f100(x,y,z)) |
(49) |
f101(x,y,g1(z)) |
→ |
g0(f101(x,y,z)) |
(50) |
f110(x,y,g0(z)) |
→ |
g0(f110(x,y,z)) |
(51) |
f111(x,y,g1(z)) |
→ |
g0(f111(x,y,z)) |
(52) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 8
components.
-
The
1st
component contains the
pair
f#010(x,g1(y),z) |
→ |
f#010(x,y,z) |
(21) |
f#010(g0(x),y,z) |
→ |
f#010(x,y,z) |
(12) |
1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#010(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g1(x1)] |
= |
1 · x1
|
[g0(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#010(x,g1(y),z) |
→ |
f#010(x,y,z) |
(21) |
f#010(g0(x),y,z) |
→ |
f#010(x,y,z) |
(12) |
and
no rules
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
f#011(x,g1(y),z) |
→ |
f#011(x,y,z) |
(22) |
f#011(g0(x),y,z) |
→ |
f#011(x,y,z) |
(13) |
1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#011(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g1(x1)] |
= |
1 · x1
|
[g0(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#011(x,g1(y),z) |
→ |
f#011(x,y,z) |
(22) |
f#011(g0(x),y,z) |
→ |
f#011(x,y,z) |
(13) |
and
no rules
could be deleted.
1.1.1.1.2.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
f#000(x,g0(y),z) |
→ |
f#000(x,y,z) |
(19) |
f#000(g0(x),y,z) |
→ |
f#000(x,y,z) |
(10) |
1.1.1.1.3 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#000(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g0(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#000(x,g0(y),z) |
→ |
f#000(x,y,z) |
(19) |
f#000(g0(x),y,z) |
→ |
f#000(x,y,z) |
(10) |
and
no rules
could be deleted.
1.1.1.1.3.1 P is empty
There are no pairs anymore.
-
The
4th
component contains the
pair
f#001(x,g0(y),z) |
→ |
f#001(x,y,z) |
(20) |
f#001(g0(x),y,z) |
→ |
f#001(x,y,z) |
(11) |
1.1.1.1.4 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#001(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g0(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#001(x,g0(y),z) |
→ |
f#001(x,y,z) |
(20) |
f#001(g0(x),y,z) |
→ |
f#001(x,y,z) |
(11) |
and
no rules
could be deleted.
1.1.1.1.4.1 P is empty
There are no pairs anymore.
-
The
5th
component contains the
pair
f#100(x,g0(y),z) |
→ |
f#100(x,y,z) |
(23) |
f#100(g1(x),y,z) |
→ |
f#100(x,y,z) |
(14) |
1.1.1.1.5 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#100(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g0(x1)] |
= |
1 · x1
|
[g1(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#100(x,g0(y),z) |
→ |
f#100(x,y,z) |
(23) |
f#100(g1(x),y,z) |
→ |
f#100(x,y,z) |
(14) |
and
no rules
could be deleted.
1.1.1.1.5.1 P is empty
There are no pairs anymore.
-
The
6th
component contains the
pair
f#101(x,g0(y),z) |
→ |
f#101(x,y,z) |
(24) |
f#101(g1(x),y,z) |
→ |
f#101(x,y,z) |
(15) |
1.1.1.1.6 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#101(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g0(x1)] |
= |
1 · x1
|
[g1(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#101(x,g0(y),z) |
→ |
f#101(x,y,z) |
(24) |
f#101(g1(x),y,z) |
→ |
f#101(x,y,z) |
(15) |
and
no rules
could be deleted.
1.1.1.1.6.1 P is empty
There are no pairs anymore.
-
The
7th
component contains the
pair
f#110(x,g1(y),z) |
→ |
f#110(x,y,z) |
(25) |
f#110(g1(x),y,z) |
→ |
f#110(x,y,z) |
(16) |
1.1.1.1.7 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#110(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g1(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#110(x,g1(y),z) |
→ |
f#110(x,y,z) |
(25) |
f#110(g1(x),y,z) |
→ |
f#110(x,y,z) |
(16) |
and
no rules
could be deleted.
1.1.1.1.7.1 P is empty
There are no pairs anymore.
-
The
8th
component contains the
pair
f#111(x,g1(y),z) |
→ |
f#111(x,y,z) |
(26) |
f#111(g1(x),y,z) |
→ |
f#111(x,y,z) |
(17) |
1.1.1.1.8 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[f#111(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
[g1(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
f#111(x,g1(y),z) |
→ |
f#111(x,y,z) |
(26) |
f#111(g1(x),y,z) |
→ |
f#111(x,y,z) |
(17) |
and
no rules
could be deleted.
1.1.1.1.8.1 P is empty
There are no pairs anymore.